Jacobi Identity

The Poisson bracket possesses the symplectic structure $[A,B] = AB - BA.$

(Note that this algebraic structure is not commutative)

Apply this rule to the brackets:

$[A, [B,C]],$ $[B, [C,A]],$ $[C, [A,B]].$

Assemble the three brackets to prove that $[A,[B,C]] + [B, [C,A]] + [C, [A,B]] = 0.$

Solution

Following this result, we apply the same rules to the following brackets:

$\left[ A,\left[ B,C \right] \right] = \left[ A,\left[ BC-CB \right] \right] = A(BC-CB) - (BC-CB)A=ABC-ACB-BCA+CBA$

$\left[ B,\left[ C,A \right] \right] = \left[ B,\left[ CA-AC \right] \right] = B(CA-AC) - (CA-AC)B=BCA-BAC-CAB+ACB$

$\left[ C,\left[ A,B \right] \right] =\left[ C,\left[ AB-BA \right] \right] = C(AB-BA) - (AB-BA)C=CAB-CBA-ABC+BAC$

We note that the order of $A,B,C$ is significant for the bracket operator. Adding the three brackets will yield 0.

Check out my other notes at Proof, Disproof, and Derivation

#Algebra #AbstractAlgebra #SymplecticStructure #PoissonBrackets #JacobiIdentity

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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